Dimensional Analysis
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Many problems in chemistry, math, physics, and engineering involve dimensional analysis. Dimensional
analysis involves calculations where the UNITS CANCEL ONE ANOTHER OUT, leaving only the
desired unit. It is best to approach each problem the same way, starting with a given, and proceeding in
such a way that the unit in the first step cancels with the unit in the second step.
You are already familiar with dimensional analysis although you might not have realized it. For example,
if your teacher asks you to bring in 2-dozen marbles, what would you do? Chances are that you would find
some marbles and count out 24 of them. So what did you do?
What you did was dimensional analysis in your head. You knew that a dozen marbles is really just 12
marbles, and since you need 2-dozen, you multiplied 12 by 2. Easy, right!?
Let’s look at a more systematic way to do this that can be applied to a number of science and engineering
problems. The approach is to always start with the GIVEN. In our case, it is 2-dozen marbles. Then we
will convert (switch from one thing to another). To do this, multiply by a fraction (really just both
multiplication and division at the same time). In every step after the given, take the units from the previous
step (in our case dozen) and PUT IT ON THE BOTTOM.
Given: 2 doz. Next step, multiply by a fraction and make sure that the doz is on the bottom.
2 doz marbles
x
__________
doz marbles
Now we need to find a CONVERSION FACTOR . This is just something that converts dozen to
something else. The conversion factor we will use is 12 marbles / doz marbles.
2 doz marbles
x
12 marbles
doz marbles
Notice what happened. Since we have the same thing on top and bottom, the units canceled. Note that the
number 2 did not cancel, just the doz marbles. Now all we have to do is multiply 2 x 12 marbles to get 24
marbles.
There is one more “trick” that we can use, and that is to remember that we cannot add apples to oranges,
but we CAN ADD LIKE TERMS!!
Example: You are 4 miles from school heading due north at a constant velocity of 15km/hr due south.
How many meters from the school will you be (and which direction) after traveling for 45 minutes?
Solution: We know that we must add meters to meters because the question specified meters for the
answer.
4mi • 5280ft • 12in • 2.54cm •
1m
–15km • 1000m • 1hr • 45min
–4812.6m
1mi
1ft
1in
100cm
1hr
1km 60min
+
mi
ft
in
cm
6437.4m
=
km/hr
m/hr
m/min -11250m
4812m S
Lets look at some ridiculous examples. Lets say that there are 22 boleighs in every 3 siefs. There are also
2 runts in every 13 siefs and 6 runts in every yaunt. So our conversion factors are:
22 boleighs 2 runts
6 runts Or, just the same
3 siefs
13 siefs
1 yaunt
3 siefs.
13 siefs
1 yaunt
22 boleighs 2 runts
6 runts
Say we want to know how many yaunts are in 3300 boleighs. Lets start with this as the given and use our
conversion factors making sure that the unit from the first step is in the bottom of the next.
3300 boleighs
x
3 siefs
22 boleighs
x
2 runts x
13 siefs
1 yaunt
6 runts
Dimensional Analysis
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Now lets make sure that all of the units cancel and leave us with what we are looking for.
3300 boleighs
x
3 siefs
22 boleighs
x
2 runts x
13 siefs
1 yaunt = 11.54 yaunts
6 runts
If there is more than one given: make sure the unit you are looking for ends up in the numerator. If not,
just invert (“Flip”) your answer.
Practice (show work and cancel units)
Name:______________
For all of the following, use the format shown showing that units cancel when one is on top and another on
bottom:
2.345km
1000m
1000mm
1cm
1in
1ft
1mi
Conversions Given
1km
1m
10mm
2.54cm
12in
5280ft
Value and
unit for this
step
2.345km
2345m 2345000mm 234500cm 92322.8in 7693.6
1.457mi
Look up the conversion factors needed and work each of the following. Round to the 3rd decimal.
1.
How many yaunts are in 4500 runts?
2.
How many seifs?
3.
How many boleighs?
4.
How many boleighs are in 99 runts?
5.
How many runts are in 99 boleighs?
6.
How many yaunts are in 2110 siefs?
7.
How many siefs are in 2110 yaunts?
8.
How many siefs are in 90 runts?
9.
How many cups are in 12 quarts?
10. How many pints?
11. How many gallons are in 112 cups?
12. pints?
13. quarts?
14. oz.?
The specific heat of liquid water is 4.184 J/gOC.
Dimensional Analysis
15. How many Joules (how much energy) are required to heat 4184g of water 20OC?
16. If you can supply 3000 J to water at 40OC, how many grams could you heat to 70OC?
Review Questions
For any given dimensional analysis problem:
1
What do we always start with?
2
What do we do in each subsequent step?
3
When do we know when to stop?
4
What should happen to the units? How?
5
What units will the answer have?
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Dimensional Analysis
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Answers:
1. 750 runts
2. 29250 seifs
3. 214,500 boleighs
4. 4719 boleighs
5. 2.08 runts
6. 54.1 yaunts
7. 82290 yaunts
8. 585 seifs
9. 48 cups
10. 24 pints
11. 7 gal
12. 56 pints
13. 28 qt
14. 896 oz
15. 4184g x 20OC x 4.184J/g OC = 350,117J
16. You need to first realize that it is heated 30OC.
To end up with g on top, you need to invert the heat capacity
1gOC/4.184J it can now be seen that to cancel the J and the OC that you will need a J on top and a OC
on the bottom: 1gOC/4.184J x 3000J/30 OC = 23.9g